How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola x^2/81-y^2/49=1?

Nov 20, 2016

The vertices are $\left(9 , 0\right)$ and $\left(- 9 , 0\right)$
The foci are F$\left(\sqrt{130} , 0\right)$ and F'$\left(- \sqrt{130} , 0\right)$
The asymptotes are $y = \frac{7}{9} x$ and $y = - \frac{7}{9} x$

Explanation:

Let's compare this equation to the gerenal equation of a left-right hyperbola

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

${x}^{2} / 81 - {y}^{2} / 49 = 1$

The center is $\left(h , k\right) = \left(0 , 0\right)$

The vertices are $\left(h \pm a , k\right) = \left(\pm 9 , 0\right)$

The slope of the asymptotes are $\pm \frac{b}{a} = \pm \frac{7}{9}$

The equations of the asymptotes are $y = \frac{b}{a} \left(x - h\right)$

The asymptotes are $y = \frac{7}{9} x$ and $y = - \frac{7}{9} x$

To determine the foci, we need $c = \pm \sqrt{{a}^{2} + {b}^{2}}$

$c = \pm \sqrt{81 + 49} = \sqrt{130}$

The foci are F$= \left(h + c , k\right) = \left(\sqrt{130} , 0\right)$

and F'$= \left(h - c , k\right) = \left(- \sqrt{130} , 0\right)$

graph{(x^2/81-y^2/49-1)(y-7/9x)(y+7/9x)=0 [-20.27, 20.29, -10.14, 10.13]}